Bribing in First-Price Auctions: Corrigendum

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, henceforth R13, studies the following game.Two ex ante identical players are about to participate in an independent-private-value first-price, sealed bid auction for one indivisible object.After the risk-neutral players learn their valuations but prior to the actual auction, player 1 can offer a take-it-or-leave-it (TIOLI) bribe to his opponent in exchange for the opponent dropping out of the contest.If the offer is accepted, player 1 is the only bidder and obtains the item for free; otherwise, both players compete non-cooperatively in the auction as usual.This is called the first-price TIOLI game. 1 R13 shows that under the restriction to continuous and monotonic bribing strategies for player 1, any equilibrium of this game must be trivial-the equilibrium bribing function employed by player 1, if it is continuous and non-decreasing, must be identically zero.In this note, we clarify the sufficient conditions under which a trivial equilibrium exists.These are less stringent than originally proposed.
Let F denote the cumulative distribution function of players' types (valuations).F is atomless, has full support on [0, 1], and its density is f .The following is Theorem 2 from R13.
Theorem 2. Suppose that F is differentiable and that it satisfies 2F (t) + tf (t) ≥ 1 for all t ∈ (0, 1], where f = F ′ .Then, the first-price TIOLI game has a trivial equilibrium.
An unfortunate fact regarding this theorem is that it is vacuously true.
However, this implies that for t ∈ (0, t), Fortunately, Theorem 2's conclusion is true under a relatively weak alternative condition.All that is required is that F is concave.The intuition is that when there is a high probability that bidders have low valuations, player 1 does not find it worthwhile to bribe player 2. This is the same intuition as initially proposed by R13.
Theorem 2 ′ .If F is concave, the first-price TIOLI game has a trivial equilibrium.
To prove this theorem we first establish a useful lemma using a geometric argument.
Lemma 1. Suppose b and x are two positive numbers such that b + x ≤ 1.Then (1) Proof.We consider two cases.In case 1, suppose 0 ≤ x ≤ b.We make our argument with reference to Figure 1a.In the figure, Hence, X ≥ D ≥ C as required.
For case 2, suppose 0 ≤ b ≤ x.The situation is as in Figure 1b.Again it is sufficient to show that D ≤ X.This inequality follows since In this case, player 1 is prescribed his optimal bid in this post-rejection-of-b information set (it is easy to show that such a best-response exists).
It is sufficient to verify that player 1 does not have a profitable deviation to a strictly positive bribe.Let b > 0 be the bribe offered by player 1 and let x be player 1's bid in the auction following the (possible) rejection of b by player 2. Obviously, we can assume that x ≤ 1 − b. 3 Given the prescribed (off-equilibrium path) behavior of player 2, the expected payoff of player 1 is On the equilibrium path, the expected payoff of bidder 1 of type θ 1 is π(θ 1 ) = θ 1 0 F (t)dt.It is sufficient to verify that for all θ 1 and for all 0 < b ≤ θ 1 and 0 ≤ x ≤ 1 − b, it is the case that Π(b, x|θ 1 ) ≤ π(θ 1 ).Let ψ(θ 1 ) ≡ π(θ 1 ) − Π(b, x|θ 1 ).Note that ψ ′ (θ 1 ) = F (θ

b+x 0 F
(t)dt is the region below the thick curve, F (t), to the left of b + x.The left-hand side of (1), F (b)x + [F (b + x) − F (b)] b, equals the shaded region, or A + B + C + D. Since F is concave and therefore F (x) ≥ F (b + x) − F (b), it easily follows that Y ≥ B. Thus, it is sufficient to show that X ≥ C. By concavity of F , D ≥ C. Finally,

Figure 1 :
Figure 1: The geometric argument in Lemma 1. Figures not to scale.